Way to evaluate this algebraic expression by hand without going insane?

EDIT: fixed asymmetrical denominator!

I was following along with a proof of Routh's theorem, and the final expression for the area of the enclosed triangle is $$1 - \frac{x}{xz + x + 1} - \frac{y}{xy + y + 1} - \frac{z}{yz + z + 1}$$ (or something extremely similar).

It's supposed to simplify to $$\frac{(xyz - 1)^2}{(xz + x + 1)(xy + y + 1)(yz + z + 1)}$$

However, I've tried doing the simplification by hand, but to no avail.

Using the WolframAlpha[...] function in Mathematica and then doing a step-by-step solution gives a few pages' worth of expansions that are obviously beyond a mortal.

Is there any way to do this that I'm missing?

• "or something extremely similar"??? Please check your expressions (both of them) since you give them out of context and don't explain where they come from - and they don't have any obvious symmetry properties (as stated) between $x,y,z$ which would be expected for a triangle. Setting $x=y=0$ gives $1-z=1$, which is false unless $z=0$ and you give no reason why that should be so. – Mark Bennet Sep 4 '14 at 18:25
• I'm guessing the last fraction is actually $$\frac z{yz+z+1}$$ to make for full symmetry in the expression. – abiessu Sep 4 '14 at 19:08
• @MarkBennet yes, there was a typo in the original question, I've fixed it. – Soham Chowdhury Sep 5 '14 at 1:43

It is easy if you fix your starting expression. The last term is $\frac{z}{yz+z+1}$ you changed that $z$ in the denominator to $y$, making it unsymmetrical.

Setting your two expressions equal to each other and mutliplying by the denominators and simplifying I get:

$$(y-z)(1+ x z + x^2 y z) = 0$$

Which means in general they are not equal. Or so it would seem.

Going off the idea of a typo: if you change the $(y z + y + 1)$ to $( y z + z + 1)$ then it simplifies to the stated answer (with the same change).
Starting with $$1 - \frac{x}{xz + x + 1} - \frac{y}{xy + y + 1} - \frac{z}{yz + y + 1}$$ I arrived to $$\frac{x^2 y^2 z^2+x^2 y^2 z-x^2 y z^2-x y z-x z^2+y-z+1}{(x y+y+1) (x z+x+1) (y z+y+1)}$$ Probably typo's somewhere.
• Yeah, the denominator of the last fraction should be $yz + z + 1$. – Soham Chowdhury Sep 5 '14 at 1:44